codeforces290E

本文详细介绍了如何通过构建奇偶图并求解输出路径问题,使用了多种数据结构如邻接表和队列,以及算法如深度优先搜索和广度优先搜索。通过实例演示了算法的应用,包括如何利用图的性质来解决实际问题。

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奇偶建图加输出路径

#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <ctime>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 0x3f3f3f3f
#define inf -0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define mem0(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,-1,sizeof(a))
#define mem(a, b) memset(a, b, sizeof(a))
typedef long long ll;
int prime[20101];
const int maxn=2001;
int a[maxn];
int Vis[maxn];
struct node{
    int x;
    int id;
}odd[maxn],even[maxn];
vector<int>ans[maxn];
struct Edge{
    int from,to,cap,flow;
    Edge(int u,int v,int c,int f):from(u),to(v),cap(c),flow(f){}
};

struct Dinic{
    int n,m,s,t;                        //结点数,边数(包括反相弧),源点编号和汇点编号
    vector<Edge>edges;                  //边表,edges[e]和edges[e^1]互为反相弧
    vector<int>G[maxn];                 //邻接表,G[i][j]表示结点i的第j条边在e数组中的序号
    bool vis[maxn];                     //BFS使用
    int d[maxn];                        //从起点到i的距离
    int cur[maxn];                      //当前弧下标
    void AddEdge(int from,int to,int cap){
        edges.push_back(Edge(from,to,cap,0));
        edges.push_back(Edge(to,from,0,0));
        m=edges.size();
        G[from].push_back(m-2);
        G[to].push_back(m-1);
    }

    void init(int n){
        this->n=n;
        for(int i=0;i<=n;i++)
            G[i].clear();
        edges.clear();
    }

    bool BFS(){
        mem0(vis);
        queue<int>Q;
        Q.push(s);
        d[s]=0;
        vis[s]=1;
        while(!Q.empty()){
            int x=Q.front();
            Q.pop();
            for(int i=0;i<G[x].size();i++){                 //只考虑残量网络中的弧
                Edge& e=edges[G[x][i]];
                if(!vis[e.to]&&e.cap>e.flow){
                    vis[e.to]=1;
                    d[e.to]=d[x]+1;
                    Q.push(e.to);
                }
            }
        }
        return vis[t];
    }

    int DFS(int x,int a){
        if(x==t||a==0)
            return a;
        int flow=0,f;
        for(int& i=cur[x];i<G[x].size();i++){
            Edge& e=edges[G[x][i]];
            if(d[x]+1==d[e.to]&&(f=DFS(e.to,min(a,e.cap-e.flow)))>0){
                e.flow+=f;
                edges[G[x][i]^1].flow-=f;
                flow+=f;
                a-=f;
                if(a==0)
                    break;
            }
        }
        return flow;
    }

    int Maxflow(int s,int t){
        this->s=s;
        this->t=t;
        int flow=0;
        while(BFS()){
            mem0(cur);
            flow+=DFS(s,INF);
        }
        return flow;
    }

    void dfs1(int u,int top){
        Vis[u]=1;
        ans[top].push_back(u);
        //printf("%d\n",G[u].size());
        for(int i=0;i<G[u].size();i++){
            Edge& e=edges[G[u][i]];
            if(Vis[e.to]||e.to==0||e.to==n+1)
                continue;
            if(a[u]&1){
                if(e.cap-e.flow==0){
                    dfs1(e.to,top);
                }
            }
            else{
                if(e.cap-e.flow!=0){
                    dfs1(e.to,top);
                }
            }
        }
    }
};
Dinic solve;

int main(){
    int n;
    mem0(prime);
    for(int i=2;i<=20001;i++){
        if(!prime[i])
        for(int j=i*i;j<=20001;j+=i){
            prime[j]=1;
        }
    }
    /*for(int i=2;i<=100;i++){
        if(prime[i]==0)
            cout<<i<<endl;
    }*/
    while(scanf("%d",&n)!=EOF){
        solve.init(n);
        for(int i=0;i<n;i++)
            ans[i].clear();
        int cnt1=0,cnt2=0;
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            if(a[i]%2==0){
                even[cnt1].x=a[i];
                even[cnt1++].id=i;
                solve.AddEdge(i,n+1,2);
            }
            else{
                odd[cnt2].x=a[i];
                odd[cnt2++].id=i;
                solve.AddEdge(0,i,2);
            }
        }
        if(n%2==1){
            printf("Impossible\n");
            continue;
        }
        for(int i=0;i<cnt2;i++)
            for(int j=0;j<cnt1;j++){
                if(prime[odd[i].x+even[j].x]==0){
                    solve.AddEdge(odd[i].id,even[j].id,1);
                }
            }
        int flow=solve.Maxflow(0,n+1);
        if(flow==n){
            mem0(Vis);
            int top=0;
            for(int i=1;i<=n;i++){
                if(Vis[i]==0){
                    solve.dfs1(i,top++);
                }
            }
            printf("%d\n",top);
            for(int i=0;i<top;i++){
                printf("%d",ans[i].size());
                for(int j=0;j<ans[i].size();j++){
                    printf(" %d",ans[i][j]);
                }
                printf("\n");
            }
        }
        else
            printf("Impossible\n");
    }
    return 0;
}
### Codeforces 887E Problem Solution and Discussion The problem **887E - The Great Game** on Codeforces involves a strategic game between two players who take turns to perform operations under specific rules. To tackle this challenge effectively, understanding both dynamic programming (DP) techniques and bitwise manipulation is crucial. #### Dynamic Programming Approach One effective method to approach this problem utilizes DP with memoization. By defining `dp[i][j]` as the optimal result when starting from state `(i,j)` where `i` represents current position and `j` indicates some status flag related to previous moves: ```cpp #include <bits/stdc++.h> using namespace std; const int MAXN = ...; // Define based on constraints int dp[MAXN][2]; // Function to calculate minimum steps using top-down DP int minSteps(int pos, bool prevMoveType) { if (pos >= N) return 0; if (dp[pos][prevMoveType] != -1) return dp[pos][prevMoveType]; int res = INT_MAX; // Try all possible next positions and update 'res' for (...) { /* Logic here */ } dp[pos][prevMoveType] = res; return res; } ``` This code snippet outlines how one might structure a solution involving recursive calls combined with caching results through an array named `dp`. #### Bitwise Operations Insight Another critical aspect lies within efficiently handling large integers via bitwise operators instead of arithmetic ones whenever applicable. This optimization can significantly reduce computation time especially given tight limits often found in competitive coding challenges like those hosted by platforms such as Codeforces[^1]. For detailed discussions about similar problems or more insights into solving strategies specifically tailored towards contest preparation, visiting forums dedicated to algorithmic contests would be beneficial. Websites associated directly with Codeforces offer rich resources including editorials written after each round which provide comprehensive explanations alongside alternative approaches taken by successful contestants during live events. --related questions-- 1. What are common pitfalls encountered while implementing dynamic programming solutions? 2. How does bit manipulation improve performance in algorithms dealing with integer values? 3. Can you recommend any online communities focused on discussing competitive programming tactics? 4. Are there particular patterns that frequently appear across different levels of difficulty within Codeforces contests?
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